Problem: The lifespans of bears in a particular zoo are normally distributed. The average bear lives $45.2$ years; the standard deviation is $10.6$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a bear living between $24$ and $34.6$ years.
Answer: $45.2$ $34.6$ $55.8$ $24$ $66.4$ $13.4$ $77$ $95\%$ $68\%$ $13.5\%$ $13.5\%$ We know the lifespans are normally distributed with an average lifespan of $45.2$ years. We know the standard deviation is $10.6$ years, so one standard deviation below the mean is $34.6$ years and one standard deviation above the mean is $55.8$ years. Two standard deviations below the mean is $24$ years and two standard deviations above the mean is $66.4$ years. Three standard deviations below the mean is $13.4$ years and three standard deviations above the mean is $77$ years. We are interested in the probability of a bear living between $24$ and $34.6$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $95\%$ of the bears will have lifespans within 2 standard deviations of the average lifespan. It also tells us that $68\%$ of the bears will have lifespans within 1 standard deviation of the mean. The probability of a particular bear living between $24$ and $34.6$ years is $\color{orange}{13.5\%}$.